Question

Let triangle ABC have vertices on a circle. Let AD be the altitude and AP be the diameter of the circle. If $\mathrm{\angle }ABC={84}^{\circ }$ and $\mathrm{\angle }BCA={60}^{\circ }$ then $\mathrm{\angle }DAP$equals to $$$$

A.$6^{\circ }$$$$$
B.${12}^{\circ }$$$$$
C.${18}^{\circ }$$$$$
D.${24}^{\circ }$$$$$

Answer 1

Hint: We join BP and CP. We use the theorem that the sum of the angles in a triangle is ${180}^{\circ }$ in triangles ABD, ACD to get $\mathrm{\angle }BAD=6^{\circ },\mathrm{\angle }CAD={30}^{\circ }$. We use the theorem that an arc of semi-circular length always subtends right angle of measure ${90}^{\circ }$ to get $\mathrm{\angle }BCP={30}^{\circ }$. We use the theorem that angle subtended by the same arc have equal measures to get $\mathrm{\angle }BAP={30}^{\circ }$ and then we find the required angle $\mathrm{\angle }DAP=\mathrm{\angle }BAP-\mathrm{\angle }BAD$.$$$$

Complete step-by-step solution
We are given in the figure a triangle ABC whose vertices are on a circle. AD is the altitude dropped on the side BC and AP is the diameter of the circle. We are further given the measure of angles $\mathrm{\angle }ABC={84}^{\circ }$ and $\mathrm{\angle }BCA={60}^{\circ }$. We are asked to find a measure of $\mathrm{\angle }DAP$. Let us join BP and CP and have the figure below.$$$$

Since AD is the altitude we have $\mathrm{\angle }ADB=\mathrm{\angle }ADC={90}^{\circ }$. We know that sum of the angles in a triangle is ${180}^{\circ }$. So we have in triangle ADB
$$\begin{array}{rl}& \Rightarrow \mathrm{\angle }ADB+\mathrm{\angle }ABD+\mathrm{\angle }BAD={180}^{\circ }\\ & \Rightarrow {90}^{\circ }+\mathrm{\angle }ABC+\mathrm{\angle }BAD={180}^{\circ }\\ & \Rightarrow {90}^{\circ }+{84}^{\circ }+\mathrm{\angle }BAD={180}^{\circ }(\because \text{ given }\mathrm{\angle }ABC={84}^{\circ })\\ & \Rightarrow {174}^{\circ }+\mathrm{\angle }BAD={180}^{\circ }\\ & \Rightarrow \mathrm{\angle }BAD=6^{\circ }\end{array}$$
We also have in triangle ADC,
$$\begin{array}{rl}& \Rightarrow \mathrm{\angle }ADC+\mathrm{\angle }ACD+\mathrm{\angle }CAD={180}^{\circ }\\ & \Rightarrow {90}^{\circ }+{60}^{\circ }+\mathrm{\angle }CAD={180}^{\circ }(\because \mathrm{\angle }ACD=\mathrm{\angle }ACB={60}^{\circ })\\ & \Rightarrow {150}^{\circ }+\mathrm{\angle }CAD={180}^{\circ }\\ & \Rightarrow \mathrm{\angle }CAD={30}^{\circ }\end{array}$$
We are given that AP is a diameter which means AP divides the circle into two semi-circles. We know that an arc of semi-circular length always subtends the right angle of measure${90}^{\circ }$ at any point on the circle. Here the semicircle has subtended $\mathrm{\angle }ABP$ and $\mathrm{\angle }ACP$ on the points $B,C$ on the circle respectively. So we have,
$$\Rightarrow \mathrm{\angle }ABP=\mathrm{\angle }ACP={90}^{\circ }$$
Then we have
$$\Rightarrow \mathrm{\angle }BCP=\mathrm{\angle }ACP-\mathrm{\angle }ACB={90}^{\circ }-{60}^{\circ }={30}^{\circ }$$
We know that angles subtended by the same arc have equal measures. Here in the circle arc BP subtends angles $\mathrm{\angle }BAP,\mathrm{\angle }BCP$. So we have,
$$\mathrm{\angle }BAP=\mathrm{\angle }BCP={30}^{\circ }$$
Then we have
$\mathrm{\angle }DAP=\mathrm{\angle }BAP-\mathrm{\angle }BAD={30}^{\circ }-6^{\circ }={24}^{\circ }$
So the correct option is D.

Note: We can alternatively solve using the sum of the angles in triangle is ${180}^{\circ }$ to get in triangle$\mathrm{\angle }BAC={36}^{\circ }$. We then join OC and use the theorem that the central angle of an arc is twice the angle subtended on appointing on the circle but not on the arc for the arc AC to get $\mathrm{\angle }OAC=6^{\circ }$ and then the required angle $\mathrm{\angle }DAP=\mathrm{\angle }CAD-\mathrm{\angle }OAC$.