The angle of elevation of the summit of a mountain from a point on the ground is $45^{\circ}$. After climbing up one km towards the summit at an inclination of $30^{\circ}$ from the ground, the angle of elevation of the summit is found to be $60^{\circ}$. Then the height (in km) of the summit from the ground is:
(1) $\frac{\sqrt{3}-1}{\sqrt{3}+1}$
(2) $\frac{\sqrt{3}+1}{\sqrt{3}-1}$
(3) $\frac{1}{\sqrt{3}-1}$
(4) $\frac{1}{\sqrt{3}+1}$

Two vertical poles are 150 m apart and the height of one is three times that of the other. If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in meters) is:
(1) 25
(2) 30
(3) $20 \sqrt { 3 }$
(4) $25 \sqrt { 3 }$

If in a triangle $A B C , A B = 5$ units, $\angle B = \cos ^ { - 1 } \left( \frac { 3 } { 5 } \right)$ and radius of circumcircle of $\triangle A B C$ is 5 units, then the area (in sq. units) of $\triangle A B C$ is:
(1) $10 + 6 \sqrt { 2 }$
(2) $8 + 2 \sqrt { 2 }$
(3) $6 + 8 \sqrt { 3 }$
(4) $4 + 2 \sqrt { 3 }$

Let in a right angled triangle, the smallest angle be $\theta$. If a triangle formed by taking the reciprocal of its sides is also a right angled triangle, then $\sin \theta$ is equal to:
(1) $\frac { \sqrt { 5 } + 1 } { 4 }$
(2) $\frac { \sqrt { 5 } - 1 } { 2 }$
(3) $\frac { \sqrt { 2 } - 1 } { 2 }$
(4) $\frac { \sqrt { 5 } - 1 } { 4 }$

A vertical pole fixed to the horizontal ground is divided in the ratio 3 : 7 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground 18 m away from the base of the pole, then the height of the pole (in meters) is :
(1) $8 \sqrt { 10 }$
(2) $6 \sqrt { 10 }$
(3) $12 \sqrt { 10 }$
(4) $12 \sqrt { 15 }$

A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point $A$, with uniform speed. At that point, angle of depression of the boat with the man's eye is $30 ^ { \circ }$ (Ignore man's height). After sailing for 20 seconds, towards the base of the tower (which is at the level of water), the boat has reached a point $B$, where the angle of depression is $45 ^ { \circ }$. Then the time taken (in seconds) by the boat from $B$ to reach the base of the tower is :
(1) 10
(2) $10 ( \sqrt { 3 } - 1 )$
(3) $10 \sqrt { 3 }$
(4) $10 ( \sqrt { 3 } + 1 )$

Let $a , b$ and $c$ be the length of sides of a triangle $ABC$ such that $\frac { a + b } { 7 } = \frac { b + c } { 8 } = \frac { c + a } { 9 }$. If $r$ and $R$ are the radius of incircle and radius of circumcircle of the triangle $ABC$, respectively, then the value of $\frac { R } { r }$ is equal to
(1) 2
(2) $\frac { 3 } { 5 }$
(3) $\frac { 5 } { 2 }$
(4) 1

The angle of elevation of the top of a tower from a point $A$ due north of it is $\alpha$ and from a point $B$ at a distance of 9 units due west of $A$ is $\cos ^ { - 1 } \left( \frac { 3 } { \sqrt { 13 } } \right)$. If the distance of the point $B$ from the tower is 15 units, then $\cot \alpha$ is equal to
(1) $\frac { 6 } { 5 }$
(2) $\frac { 9 } { 5 }$
(3) $\frac { 4 } { 3 }$
(4) $\frac { 7 } { 3 }$

A tower $PQ$ stands on a horizontal ground with base $Q$ on the ground. The point $R$ divides the tower in two parts such that $QR = 15$ m. If from a point $A$ on the ground the angle of elevation of $R$ is $60 ^ { \circ }$ and the part $PR$ of the tower subtends an angle of $15 ^ { \circ }$ at $A$, then the height of the tower is
(1) $5(2\sqrt { 3 } + 3)$ m
(2) $5(\sqrt { 3 } + 3)$ m
(3) $10(\sqrt { 3 } + 1)$ m
(4) $10(2\sqrt { 3 } + 1)$ m

Let a vertical tower $A B$ of height $2 h$ stands on a horizontal ground. Let from a point $P$ on the ground a man can see upto height $h$ of the tower with an angle of elevation $2 \alpha$. When from $P$, he moves a distance $d$ in the direction of $\overrightarrow { A P }$, he can see the top $B$ of the tower with an angle of elevation $\alpha$. If $d = \sqrt { 7 } h$, then $\tan \alpha$ is equal to
(1) $\sqrt { 5 } - 2$
(2) $\sqrt { 3 } - 1$
(3) $\sqrt { 7 } - 2$
(4) $\sqrt { 7 } - \sqrt { 3 }$

For a triangle $ABC$, the value of $\cos 2A + \cos 2B + \cos 2C$ is least. If its inradius is 3 and incentre is $M$, then which of the following is NOT correct?
(1) Perimeter of $\triangle ABC$ is $18\sqrt{3}$
(2) $\sin 2A + \sin 2B + \sin 2C = \sin A + \sin B + \sin C$
(3) $\overrightarrow{MA} \cdot \overrightarrow{MB} = -18$
(4) area of $\triangle ABC$ is $\frac{27\sqrt{3}}{2}$

In a triangle $A B C$, if $\cos A + 2 \cos B + \cos C = 2$ and the lengths of the sides opposite to the angles $A$ and $C$ are 3 and 7 respectively, then $\cos A - \cos C$ is equal to
(1) $\frac { 9 } { 7 }$
(2) $\frac { 10 } { 7 }$
(3) $\frac { 5 } { 7 }$
(4) $\frac { 3 } { 7 }$

In a triangle $\mathrm { ABC } , \mathrm { BC } = 7 , \mathrm { AC } = 8 , \mathrm { AB } = \alpha \in \mathrm { N }$ and $\cos \mathrm { A } = \frac { 2 } { 3 }$. If $49 \cos ( 3 \mathrm { C } ) + 42 = \frac { \mathrm { m } } { \mathrm { n } }$, where $\operatorname { gcd } ( \mathrm { m } , \mathrm { n } ) = 1$, then $\mathrm { m } + \mathrm { n }$ is equal to $\_\_\_\_$

Q84. In a triangle $\mathrm { ABC } , \mathrm { BC } = 7 , \mathrm { AC } = 8 , \mathrm { AB } = \alpha \in \mathrm { N }$ and $\cos \mathrm { A } = \frac { 2 } { 3 }$. If $49 \cos ( 3 \mathrm { C } ) + 42 = \frac { \mathrm { m } } { \mathrm { n } }$, where $\operatorname { gcd } ( \mathrm { m } , \mathrm { n } ) = 1$, then $\mathrm { m } + \mathrm { n }$ is equal to $\_\_\_\_$

Q2 As shown in the figure, a triangle $ABD$ is inscribed in a semi-circle with the diameter $BC$, where
$$\mathrm{AB}=3, \quad \mathrm{BD}=5, \quad \tan\angle\mathrm{ABD}=\frac{3}{4}.$$
We are to find the lengths of the three sides $BC$, $CD$ and $DA$ of the quadrangle $ABCD$ and the area $S$ of the quadrangle $ABCD$.
First, since $\cos\angle\mathrm{ABD}=\frac{\mathbf{L}}{\mathbf{M}}$, we have $\mathrm{DA}=\sqrt{\mathbf{NO}}$.
Also, since $\sin\angle\mathrm{ABD}=\frac{\mathbf{P}}{\mathbf{Q}}$, we have $\mathrm{BC}=\frac{\square\mathbf{R}\sqrt{\mathbf{ST}}}{\mathbf{U}}$ and thus $\mathrm{CD}=\frac{\mathbf{V}}{\mathbf{W}}$. From these we obtain
$$S = \frac{\mathbf{XXY}}{\mathbf{XY}}.$$

A quadrangle $ABCD$ which is inscribed in a circle $O$ satisfies
$$\mathrm { AB } = \mathrm { BC } = \sqrt { 2 } , \quad \mathrm { BD } = \frac { 3 \sqrt { 3 } } { 2 } , \quad \angle \mathrm { ABC } = 120 ^ { \circ } ,$$
where
$$AD > CD .$$
(1) Then $\mathrm { AC } = \sqrt { \mathbf { A } }$, and the radius of circle O is $\sqrt { \mathbf { B } }$.
(2) Set $x = \mathrm { AD }$. Since $\angle \mathrm { ADB } = \mathbf { CD }^\circ$, $x$ satisfies
$$4 x ^ { 2 } - \mathbf { E } \mathbf { F } x + \mathbf { G H } = 0 .$$
Also, set $y = \mathrm { CD }$. In the same way, it follows that $y$ satisfies
$$4 y ^ { 2 } - \mathbf { I J } y + \mathbf { K } \mathbf { L } = 0 .$$
Thus, noting (1), we obtain $AD$ and $CD$.

Consider a triangle ABC where
$$\mathrm { AB } = 10 , \quad \angle \mathrm { B } = 30 ^ { \circ }$$
and the radius of its inscribed circle O is 1.
(1) Set $a = \mathrm { BC }$ and $b = \mathrm { CA }$. Finding the area $S$ of the triangle ABC by two different methods, we have
$$S = \frac { \mathbf { A } } { \mathbf{B} } \mathbf { B } ,$$
and
$$S = \frac { \mathbf { C } } { \mathbf{C} } \mathbf { D } ( a + b + \mathbf { E C } ) .$$
Hence we obtain
$$b = \mathbf { G } a - \mathbf { H I } .$$
Since the relationship between $a$ and $b$ can also be expressed by the equation
$$b ^ { 2 } = a ^ { 2 } - \mathbf { J K } \sqrt { \mathbf { L } } a + \mathbf { M N O } ,$$
we have
$$a = \frac { \mathbf { P Q } - \mathbf { R } } { 3 } \sqrt { \mathbf { S } } , \quad b = \frac { \mathbf { T U } } { \mathbf{P} } \cdot \frac { \mathbf { V } } { 3 } \sqrt { \mathbf { W } } .$$
(2) Let D denote the point of intersection of the segment BC and the straight line which passes through the two points A and O. We denote the area of the triangle OBC by $S ^ { \prime }$. Since
$$S : S ^ { \prime } = \mathbf { X } : 1 ,$$
it follows that
$$\mathrm { AO } : \mathrm { OD } = \mathbf { Y } : 1 .$$

Consider a triangle ABC where $\mathrm{AB} = 8$, $\mathrm{AC} = 5$ and $\angle\mathrm{BAC} = 60^\circ$. Take point D and point E on sides AB and AC, respectively, such that the segment DE divides triangle ABC into two parts with the equal areas. Set $x = \mathrm{AD}$.
(1) When we represent AE in terms of $x$, we have $\mathrm{AE} = \frac{\mathbf{JK}}{x}$.
(2) When E moves along side AC, the range of the values which $x$ can take is
$$\mathbf{L} \leq x \leqq \mathbf{M}.$$
(3) Since $\mathrm{DE}^2 = \left(x - \frac{\mathbf{NO}}{x}\right)^2 + \mathbf{PQ}$, the length of segment DE is minimized at $x = \mathbf{R}\sqrt{\mathbf{S}}$, and its length there is $\mathbf{T}$.$\mathbf{U}$.

Suppose that in the figure to the right
$$\mathrm { AB } = 4 , \quad \mathrm { AC } = 5 , \quad \cos \angle \mathrm { BAC } = \frac { 1 } { 8 }$$
and
$$\angle \mathrm { BAD } = \angle \mathrm { ACB } , \quad \angle \mathrm { CAE } = \angle \mathrm { ABC } .$$
(1) When we denote the area of $\triangle \mathrm { ABC }$ by $S$, we have
$$S = \frac { \square \mathbf { A B } \sqrt { \square \mathbf { C } } } { \square } .$$
Also $\mathrm { BC } = \mathbf { E }$.
(2) Furthermore, when we denote the areas of $\triangle \mathrm { ABD }$ and $\triangle \mathrm { ACE }$ by $S _ { 1 }$ and $S _ { 2 }$, respectively, we have
$$S : S _ { 1 } : S _ { 2 } = 1 : \frac { \mathbf { F } } { \mathbf{G} } : \frac { \mathbf { H I } } { \mathbf { J } } .$$
(3) When we denote the area of $\triangle \mathrm { ADE }$ by $T$, we have
$$T = \frac { \mathbf { L M } \sqrt { \mathbf { N } } } { \mathbf { O P } } .$$
Also $\mathrm { DE } = \dfrac { \mathbf { Q } } { \mathbf{R} }$.

A triangle ABC satisfies
$$\mathrm{AB} = 7, \quad \mathrm{BC} = 8, \quad \mathrm{CA} = 6.$$
We denote the center and the radius of the circumscribed circle of this triangle ABC by O and $r$, respectively. We draw two straight lines which are tangent to this circumscribed circle at the points B and C, and denote the point of intersection of these straight lines by D.
We see that
$$\cos\angle\mathrm{BAC} = \frac{\mathbf{A}}{\mathbf{B}}, \quad \sin\angle\mathrm{BAC} = \frac{\sqrt{\mathbf{CD}}}{\mathbf{E}},$$
$$r = \frac{\mathbf{FG}\sqrt{\mathbf{HI}}}{\mathbf{JK}}, \quad \mathrm{BD} = \mathbf{LM}.$$
Furthermore, if P is a point on the circumscribed circle, the shortest possible length of the segment DP is $\frac{\mathbf{NO}\sqrt{\mathbf{PQ}}}{\mathbf{R}}$.

Suppose that a quadrangle ABCD which is inscribed in a circle has the side lengths
$$\mathrm { AB } = \sqrt { 2 } , \quad \mathrm { BC } = \mathrm { CD } = 2 , \quad \mathrm { DA } = \sqrt { 6 } .$$
(1) Let us set $\theta = \angle \mathrm { BAD }$. We have the two equalities
$$\begin{aligned} & \mathrm { BD } ^ { 2 } = \mathbf { A } - \mathbf { B } \sqrt { \mathbf { C } } \cos \theta , \\ & \mathrm { BD } ^ { 2 } = \mathbf { D } + \mathbf { E } \cos \theta . \end{aligned}$$
Hence,
$$\theta = \mathbf { F G } { } ^ { \circ } , \quad \mathrm { BD } = \mathbf { H } \text {. } \mathbf { I } \text {. }$$
(2) Furthermore, we have
$$\angle \mathrm { BAC } = \mathbf { J K } ^ { \circ } , \quad \angle \mathrm { BCA } = \mathbf { L M } ^ { \circ } \text { and } \mathrm { AC } = \mathbf { N } + \sqrt { \mathbf { O } } \text {. }$$
We also have
$$\sin \angle \mathrm { ADC } = \frac { \sqrt { \mathbf { P } } } { \mathbf{Q} } ( \sqrt { \mathbf { R } } + \mathbf { S } )$$
(3) Let us denote the point of intersection of the straight line AD and the straight line BC by E. We have $\mathrm { EB } = \mathbf { T } + \mathbf { U } \sqrt { \mathbf { V } }$.

We have a triangle which has sides of the lengths 15, 19 and 23. We make it into an obtuse triangle by shortening each of its sides by $x$. What is the range of values that $x$ can take?
First, since $15 - x$, $19 - x$ and $23 - x$ can be the lengths of the sides of a triangle, it follows that $$x < \mathbf{AB}.$$
In addition, such a triangle is an obtuse triangle only when $x$ satisfies $$x^2 - \mathbf{CD}x + \mathbf{EF} < 0.$$
By solving this quadratic inequality, we have $$\mathbf{G} < x < \overline{\mathbf{HI}}.$$
Hence, the range of $x$ is $$\mathbf{J} < x < \mathbf{KL}.$$

In the figure to the right, let $$\mathrm{OA} = 6, \quad \mathrm{OB} = 3, \quad \angle\mathrm{AOB} = 120^\circ,$$ and let the point Q denote the point of intersection of the bisector of $\angle\mathrm{XAB}$ and the bisector of $\angle\mathrm{ABY}$. Let P denote the point of intersection of segment AB and segment OQ. We are to find the length of segment PQ.
(1) First of all, we see that $\mathrm{AB} = \square\sqrt{\mathbf{A}}$ and that the area of triangle OAB is $\mathbf{BC}\sqrt{\mathbf{D}}$.
(2) For $\mathbf{F}$ and $\mathbf{G}$ in the following, choose the correct answer from among choices (0) $\sim$ (4), just below. (0) AB (1) AP (2) AQ (3) BP (4) BQ
Since AQ is the bisector of the exterior angle of $\angle\mathrm{A}$ of triangle OAP and BQ is the bisector of the exterior angle of $\angle\mathrm{B}$ of triangle OBP, we have $$\begin{aligned} \mathrm{OQ} : \mathrm{PQ} &= \mathrm{OA} : \mathbf{F} \\ &= \mathrm{OB} : \mathbf{G} \end{aligned}$$
(3) Thus we see $\mathrm{AP} = \mathbf{H}\sqrt{\mathbf{I}}$. Since $\angle\mathrm{AOP} = \mathbf{JK}^\circ$, we have $\mathrm{OP} = \square\mathbf{L}$. Hence we have $\mathrm{PQ} = \mathbf{M} + \mathbf{N}\sqrt{\mathbf{O}}$.

The figure to the right is a net for the tetrahedron OABC. This tetrahedron satisfies
$$\begin{gathered} \mathrm { BC } = 10 , \quad \mathrm { AC } = 8 , \quad \sin \angle \mathrm { ACB } = \frac { 3 } { 4 } , \\ \mathrm { OA } = 4 , \quad \triangle \mathrm { ABC } \equiv \triangle \mathrm { OBC } . \end{gathered}$$
(1) The area of the triangle ABC is $\mathbf { A B }$.
(2) Let AH denote the perpendicular line drawn from point A to side BC. The length of AH is $\mathbf { C }$.
(3) Let $\theta$ denote the angle formed by the plane ABC and the plane OBC. Then we have
$$\cos \theta = \frac { \mathbf { D } } { \mathbf { E } } , \quad \sin \theta = \frac { \mathbf { F } \sqrt { \mathbf { G } } } { \mathbf { H } } .$$
(4) The volume of the tetrahedron OABC is $\frac { \mathbf { IJ } \sqrt { \mathbf { K } } } { \mathbf { L } }$.

The triangle $ABC$ satisfies
$$\mathrm { AB } = 2 , \quad \mathrm { BC } = 3 , \quad \mathrm { CA } = 4 .$$
(1) When we set $\angle \mathrm { ABC } = \theta$, the inner product $\overrightarrow { \mathrm { AB } } \cdot \overrightarrow { \mathrm { BC } }$ of the vectors $\overrightarrow { \mathrm { AB } }$ and $\overrightarrow { \mathrm { BC } }$ is
$$\overrightarrow { \mathrm { AB } } \cdot \overrightarrow { \mathrm { BC } } = \mathbf { AB } \cos \theta .$$
Finding the value of $\cos \theta$ from the law of cosines, we obtain
$$\overrightarrow { \mathrm { AB } } \cdot \overrightarrow { \mathrm { BC } } = \frac { \mathbf { C } } { \mathbf { D } } .$$
(2) We divide the side BC into $n$ equal parts by the points $\mathrm { P } _ { 1 } , \mathrm { P } _ { 2 } , \cdots , \mathrm { P } _ { n - 1 }$ which are arranged in ascending order of the distance from B, and set $\mathrm { B } = \mathrm { P } _ { 0 } , \mathrm { C } = \mathrm { P } _ { n }$. We are to find the value of $\lim _ { n \rightarrow \infty } \frac { 1 } { n } \sum _ { k = 1 } ^ { n } \overrightarrow { \mathrm { AP } _ { k - 1 } } \cdot \overrightarrow { \mathrm { AP } _ { k } }$.
When we calculate the inner product of $\overrightarrow { \mathrm { AP } _ { k - 1 } }$ and $\overrightarrow { \mathrm { AP } _ { k } }$ using (1), we have
$$\overrightarrow { \mathrm { AP } _ { k - 1 } } \cdot \overrightarrow { \mathrm { AP } _ { k } } = \mathbf { E } + \frac { \mathbf { F } } { 2 n } + \mathbf { G }$$
Hence we obtain
$$\lim _ { n \rightarrow \infty } \frac { 1 } { n } \sum _ { k = 1 } ^ { n } \overrightarrow { \mathrm { AP } _ { k - 1 } } \cdot \overrightarrow { \mathrm { AP } _ { k } } = \frac { \mathbf { IJ } } { \mathbf { K } }$$